Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 2)
2.
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is:
Answer: Option
Explanation:
Let the sum be Rs. x. Then,
| C.I. = |  |
x |  |
1 + | 4 |  |
2 | - x |  |
= | |
676 | x | - x | |
= | 51 | x. |
| 100 | 625 | 625 |
| S.I. = | |
x x 4 x 2 | |
= | 2x | . |
| 100 | 25 |
 |
51x | - | 2x | = 1 |
| 625 | 25 |
x = 625.
Discussion:
149 comments Page 10 of 15.
GAGAN said:
9 years ago
If the compound interest is 104 and simple interest is 100 for the time of 2 years find the rate of interest.
Can anyone solve this problem.
Can anyone solve this problem.
Rishi said:
9 years ago
I guess X must be the principal but not the sum.
Ramesh said:
9 years ago
P = D[100/r] * 2.
P = 1[100 * 100/4 * 4].
P = 625.
P = 1[100 * 100/4 * 4].
P = 625.
Jyoti said:
9 years ago
P = D * 100^2/R^2 = 1*100^2/16 = 625.
Amit said:
9 years ago
Principal for two years = pr2/100*100.
Three years = pr2(300+r)/100*100*100.
Three years = pr2(300+r)/100*100*100.
Princess said:
9 years ago
Can anyone tell me how can we find the sum and rate when compound interest for two successive years are are given?
Priya said:
10 years ago
Should we take x as principle? Or we can take any other variable like P?
Anu said:
10 years ago
Krishna please solve last equation also.
Santhosh P said:
10 years ago
Difference for 2 years = P*(R/100)^2.
=> P*(4/100)^2 = 1.
=> P = 625.
=> P*(4/100)^2 = 1.
=> P = 625.
Sagy said:
10 years ago
I did not understand this sum please help give me a simple way to solve this sum.
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